Let \( \alpha, \beta \) be the distinct roots of the equation $$ x^2 - (t^2 - 5t + 6)x + 1 = 0, \, t \in \mathbb{R} \, \text{and} \, a_n = \alpha^n + \beta^n. $$ Then the minimum value of \( \frac{a_{2023} + a_{2025}}{a_{2024}} \) is:
- \( \frac{1}{4} \)
- \( -\frac{1}{2} \)
- \( -\frac{1}{4} \)
- \( \frac{1}{2} \)
The Correct Option is C
Approach Solution - 1
By Newton's theorem, the recurrence relation for \( a_n \) is:
\[ a_{n+2} - \left(t^2 - 5t + 6\right)a_{n+1} + a_n = 0. \]
Using this relation:
\[ a_{2025} + a_{2023} = \left(t^2 - 5t + 6\right)a_{2024}. \]
Substitute into the given expression:
\[ \frac{a_{2023} + a_{2025}}{a_{2024}} = t^2 - 5t + 6. \]
The quadratic \( t^2 - 5t + 6 \) can be expressed as:
\[ t^2 - 5t + 6 = \left(t - \frac{5}{2}\right)^2 - \frac{1}{4}. \]
The minimum value of \( \left(t - \frac{5}{2}\right)^2 \) is \( 0 \), which occurs when \( t = \frac{5}{2} \). Substituting this into the equation:
\[ \text{Minimum value} = -\frac{1}{4}. \]
Approach Solution -2
To solve the problem, we need to find the minimum value of \( \frac{a_{2023} + a_{2025}}{a_{2024}} \), where \( a_n = \alpha^n + \beta^n \) and \( \alpha, \beta \) are roots of the quadratic equation \( x^2 - (t^2 - 5t + 6)x + 1 = 0 \).
First, let's analyze the equation \( x^2 - (t^2 - 5t + 6)x + 1 = 0 \). The roots \( \alpha \) and \( \beta \) are given by the equation:
\(\alpha + \beta = t^2 - 5t + 6\)
\(\alpha \beta = 1\)
Using the property of roots of quadratic equations, the sequence \( a_n \) can be derived from the recurrence relation:
\(a_n = (t^2 - 5t + 6) a_{n-1} - a_{n-2}\)
Given initial conditions for sequence:
\(a_0 = 2\) (since \( a_0 = \alpha^0 + \beta^0 = 1 + 1 \))
\(a_1 = t^2 - 5t + 6\) (since \( a_1 = \alpha + \beta \))
The required expression is:
\(\frac{a_{2023} + a_{2025}}{a_{2024}}\)
This expression is equivalent to:
\(a_{2024} \, = \, (t^2 - 5t + 6) a_{2023} - a_{2022}\)
\(a_{2025} \, = \, (t^2 - 5t + 6) a_{2024} - a_{2023}\)
Thus, the expression becomes:
\(\frac{a_{2023} + ((t^2 - 5t + 6) a_{2024} - a_{2023})}{a_{2024}} \, = \, \frac{(t^2 - 5t + 6) a_{2024}}{a_{2024}}\)
Simplifying this, we find:
\(\frac{a_{2023} + a_{2025}}{a_{2024}} = t^2 - 5t + 6 - 2\)
Which further simplifies to:
\(t^2 - 5t + 4\)
To find the minimum value of \( t^2 - 5t + 4 \), we complete the square:
\(t^2 - 5t + 4 = (t - \frac{5}{2})^2 - \frac{9}{4}\)
The minimum value occurs at \( t = \frac{5}{2} \), with the minimum value being:
\(-\frac{9}{4}\)
Therefore, the minimum value of \( \frac{a_{2023} + a_{2025}}{a_{2024}} \) is \(-\frac{1}{4}\).
Hence, the correct answer is \(-\frac{1}{4}\).
Top JEE Main Mathematics Questions
- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- If , then the general value of is:
- The general solution of
- If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____
Top JEE Main Sequence and series Questions
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- \(3, a, b, c\) are in Ap and \(3, a-1, b+1, c+9\) are in GP. Then AM of \(a, b, c\) is
- If \( \alpha, \beta \) are the roots of the equation \( x^2 - x - 1 = 0 \) and \( S_n = 2023 \alpha^n + 2024 \beta^n \), then:
- The function \( f(x) = 2x + 3(x)^{\frac{2}{3}}, x \in \mathbb{R} \), has
- Let \( S_n \) be the sum to \( n \)-terms of an arithmetic progression \( 3, 7, 11, \ldots \).
If \( 40 < \left( \frac{6}{n(n+1)} \sum_{k=1}^{n} S_k \right) <42 \), then \( n \) equals ___.
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.